$\begin{aligned} & \text { } 1^2+\left(1^2+2^2\right)+\left(1^2+2^2+3^2\right)+\ldots \\ & +\left(1^2+2^2+\ldots+n^2\right) \\ & =\sum_{i=1}^n\left(1^2+2^2+3^2+\ldots+i^2\right)=\sum_{i=1}^n \sum_{j=1}^i j^2 \\ & =\sum_{i=1}^n \frac{i(i+1)(2 i+1)}{6}\left\{\because \Sigma n^2=\frac{n(n+1)(2 n+1)}{6}\right\} \\ & =\sum_{i=1}^n \frac{2 i^3+3 i^2+i}{6}=\frac{1}{3} \sum_{i=1}^n i^3+\frac{1}{2} \sum_{i=1}^n i^2+\frac{1}{6} \sum_{i=1}^n i \\ & =\frac{1}{3}\left[\frac{n(n+1)}{2}\right]^2+\frac{1}{2}\left[\frac{n(n+1)(2 n+1)}{6}\right]+\frac{1}{6}\left[\frac{n(n+1)}{2}\right] \\ & \left\{\begin{array}{c}\because \Sigma n^3=\left[\frac{n(n+1)}{2}\right]^2 ; \\ \Sigma n=\frac{n(n+1)}{2}\end{array}\right\} \\ & =\frac{1}{3} \cdot \frac{n^2(n+1)^2}{4}+\frac{1}{2} \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{1}{6} \cdot \frac{n(n+1)}{2} \\ & =\frac{n^2(n+1)^2}{12}+\frac{n(n+1)(2 n+1)}{12}+\frac{n(n+1)}{12} \\ & =\frac{n^2(n+1)^2+n(n+1)(2 n+1)+n(n+1)}{12} \\ & =\frac{n(n+1)[n(n+1)+(2 n+1)+1]}{12} \\ & =\frac{n(n+1)}{12}\left[n^2+n+2 n+2\right] \\ & =\frac{n(n+1)}{12}\left[n^2+3 n+2\right]=\frac{n(n+1)}{12}[(n+1)(n+2)] \\ & =\frac{n(n+1)(n+1)(n+2)}{12}=\frac{n(n+1)^2(n+2)}{12} \\ & \end{aligned}$